What is the integral from 0 to 72 of sin^2(θ)sin(2θ) ?

The integral from 0 to 72 of sin^2(θ)sin(2θ) is (5(3+sqrt(5)))/(64)

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integral from 0 to 72 of sin^2(θ)sin(2θ)

\int_{0}^{7 2^{\circ}} sin^{2} (θ) sin (2 θ) d θ

\frac{5 ( 3 + 5 )}{64}

Decimal Notation

0.40906 \dots

Solution steps

\int_{0}^{7 2^{\circ}} sin^{2} (θ) sin (2 θ) d θ

Rewrite using trig identities

= \int_{0}^{7 2^{\circ}} \frac{1}{2} sin (2 θ) (1 - cos (2 θ)) d θ

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{7 2^{\circ}} sin (2 θ) (1 - cos (2 θ)) d θ

Apply u-substitution

: \int_{0}^{1 + \frac{5 + 1}{4}} \frac{u}{2} d u

= \frac{1}{2} \cdot \int_{0}^{1 + \frac{5 + 1}{4}} \frac{u}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{0}^{1 + \frac{5 + 1}{4}} u d u

Apply the Power Rule

: [\frac{u ^{2}}{2}]_{0}^{1 + \frac{5 + 1}{4}}

= \frac{1}{2} \cdot \frac{1}{2} [\frac{u ^{2}}{2}]_{0}^{1 + \frac{5 + 1}{4}}

Simplify

\frac{1}{2} \cdot \frac{1}{2} [\frac{u ^{2}}{2}]_{0}^{1 + \frac{5 + 1}{4}} : \frac{1}{4} [\frac{u ^{2}}{2}]_{0}^{1 + \frac{5 + 1}{4}}

= \frac{1}{4} [\frac{u ^{2}}{2}]_{0}^{1 + \frac{5 + 1}{4}}

Compute the boundaries:

\frac{5 ( 3 + 5 )}{16}

= \frac{1}{4} \cdot \frac{5 ( 3 + 5 )}{16}

Simplify

= \frac{5 ( 3 + 5 )}{64}

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What is the integral from 0 to 72 of sin^2(θ)sin(2θ) ?
The integral from 0 to 72 of sin^2(θ)sin(2θ) is (5(3+sqrt(5)))/(64)